The point A (-7, 5) is reflected over the line x = -5, and then is reflected over the line x = 2. What are the coordinates of A'?
(7, 19)
(10, 5)
(7, 5)
(10, 19)
Question #45
MultipleChoice
The vertices of an isosceles triangle are A(3, 6), B(7, 2), and C(4, 3).
What is the equation of the triangle's line of symmetry?
y = x + 1
y = x - 1
y = x
y = -x + 1
Question #46
MultipleChoice
The area of a circle is 16π cm2. What is the circle's circumference?
2π cm
4π cm
8π cm
16π cm
Question #47
MultipleChoice
If AC = 20, AE = 25, and AB = 5, what is the length of ?
4
5
6
3
Question #48
MultipleChoice
A square pyramid is 6 feet on each side. The height of the pyramid is 4 feet. What is the total area of the pyramid?
60 ft2
156 ft2
96 ft2
120 ft2
Question #49
MultipleSelect
A'B'C'D' is the image of ABCD. What transformation(s) would result in this image?
ABCD is reflected over the line y = x.
ABCD is dilated about the origin with a scale factor of -1.
ABCD is reflected across the y-axis and then the x-axis.
ABCD is rotated 180° around the origin.
Question #50
MultipleChoice
A circle has a diameter with endpoints (-8, 2) and (-2, 6).
What is the equation of the circle?
r2 = (x - 3)2 + (y + 4)2
r2 = (x - 5)2 + (y + 4)2
r2 = (x + 5)2 + (y - 4)2
r2 = (x + 3)2 + (y - 4)2
Wow,
do you really expect somebody to do this whole assignment for you?
The point A (-7, 5) is reflected over the line x = -5, and then is reflected over the line x = 2. To find the coordinates of A', we follow these steps:
1. Reflect A over the line x = -5: Since A is on the left side of x = -5, A' will be on the right side of x = -5, the same distance away from the line. The x-coordinate of A' will be 2 units to the right of x = -5, which is x = -3. The y-coordinate of A' remains the same. So the coordinates of the reflected point A' are (-3, 5).
2. Reflect A' over the line x = 2: Since A' is on the left side of x = 2, A'' will be on the right side of x = 2, the same distance away from the line. The x-coordinate of A'' will be 4 units to the right of x = 2, which is x = 6. The y-coordinate of A'' remains the same. So the coordinates of the reflected point A'' are (6, 5).
Therefore, the coordinates of A' are (6, 5), so the correct answer is (7, 5).
To find the coordinates of A' after reflecting point A (-7, 5) over the line x = -5 and then over the line x = 2, we can use the concept of reflections.
Step 1: Reflect A over the line x = -5
To reflect a point over a vertical line, the y-coordinate stays the same, but the x-coordinate changes sign. So, when reflecting A (-7, 5) over x = -5, the x-coordinate changes sign to become positive, and the y-coordinate remains the same. This gives us A' (7, 5).
Step 2: Reflect A' over the line x = 2
Similar to the previous step, reflecting A' (7, 5) over x = 2 means changing the sign of the x-coordinate, while keeping the y-coordinate the same. This gives us the final coordinates A' (-7, 5).
Therefore, the coordinates of A' are (-7, 5).
As for the other questions, I will explain how to solve them one by one:
Question #45:
To find the equation of the line of symmetry of an isosceles triangle, we need to find the midpoint of the base. The side lengths of the triangle are not given, so we can assume that AB and AC are the equal sides. The midpoint of BC (the base) can be found by taking the average of the x-coordinates and y-coordinates of B(7, 2) and C(4, 3). This gives us the midpoint M(5.5, 2.5). The line passing through point M with a slope of 1 (the negative reciprocal of the slope of BC) is the line of symmetry. The equation of this line is y = x - 1.
Question #46:
The area of a circle is given by the formula A = πr^2, where r is the radius of the circle. In this question, we are given that the area is 16π cm^2. So, we can set up the equation 16π = πr^2 and solve for r. By canceling out π on both sides, we get r^2 = 16. Taking the square root of both sides, we find r = 4 cm. The circumference of a circle is given by the formula C = 2πr, so the circumference of this circle is 2π(4) = 8π cm.
Question #47:
Based on the given information, the side AB = 5, side AC = 20, and side AE = 25. We can see that AE is the longest side, so according to the Triangle Inequality Theorem, AE = AB + BE, where BE represents the length of side BE. Substituting the given values, we get 25 = 5 + BE, and solving for BE, we find that BE = 20. Therefore, the length of side BE is 20.
Question #48:
To find the total area of a square pyramid, we need to find the area of the base and the lateral faces. The base of the pyramid is a square, and each side is given as 6 feet. The formula for the area of a square is A = side^2, so the area of the base is 6^2 = 36 square feet. The lateral faces of the pyramid are triangular, and the formula for the area of a triangle is A = 1/2 * base * height. Since the base of each triangular face is the side length of the base square, which is 6 feet, and the height is the height of the pyramid, which is 4 feet, the area of each triangular face is 1/2 * 6 * 4 = 12 square feet. There are 4 triangular faces, so the total area of the lateral faces is 4 * 12 = 48 square feet. Adding the area of the base and the lateral faces, we get a total area of 36 + 48 = 84 square feet.
Question #49:
The transformation(s) that would result in the image A'B'C'D' from the original figure ABCD are: ABCD is reflected across the y-axis and then the x-axis. This would result in a 180-degree rotation of the figure. So, the correct options are:
- ABCD is reflected across the y-axis and then the x-axis.
- ABCD is rotated 180° around the origin.
Question #50:
To find the equation of a circle, we need the coordinates of its center and its radius. The midpoint of the diameter with endpoints (-8, 2) and (-2, 6) gives us the center of the circle. Using the midpoint formula, the x-coordinate of the center is (-8 + -2) / 2 = -5 and the y-coordinate is (2 + 6) / 2 = 4. So, the center of the circle is (-5, 4). The radius of the circle is half the length of the diameter, which is (distance between endpoints) / 2. Using the distance formula, the distance between the endpoints is sqrt((-8 - -2)^2 + (2 - 6)^2) = sqrt(36 + 16) = sqrt(52) = 2sqrt(13). Therefore, the radius is (2sqrt(13)) / 2 = sqrt(13). The equation of a circle with center (h, k) and radius r is (x - h)^2 + (y - k)^2 = r^2. Substituting the values, we get (x - (-5))^2 + (y - 4)^2 = (sqrt(13))^2, which simplifies to (x + 5)^2 + (y - 4)^2 = 13. So, r^2 = (x + 5)^2 + (y - 4)^2 is the equation of the circle.